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I was working in a login system session_is_registered but this function doesnt work in php5.4+ so I replaced:

   if (session_is_registered('id_user') == true && session_is_registered ('username') == true){}

to 

if (isset($_SESSION['id_user']) && isset($_SESSION['username'])){}

...and it works.

But what happens if a want to replace (becouse I need to use this expression)

if (!session_is_registered('id_user') || !session_is_registered('username')){}

Do I need to use !isset() or empty()?

S.I.
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Max
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1 Answers1

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You can use like this way

if (isset($_SESSION['id_user']) && isset($_SESSION['username'])){
      /* this will true if both are set */
}

if (!isset($_SESSION['id_user']) || !isset($_SESSION['username'])){
      /* this will true if one of them is not set */
}

isset will give true if variable is set
empty will give true if variable is set and its blank or null

isset() is NOT equivalent to empty().

Jazzzzzz
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  • thanks you, please help me with this expression `if ((!isset($_SESSION['id_user']) && empty($_SESSION['id_user']) || (!isset($_SESSION['usernamer']) && empty($_SESSION['username'])){}` would be correct? – Max Oct 23 '16 at 12:48
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    try this it's more safe to use `if (!isset($_SESSION['id_user']) || empty($_SESSION['id_user']) || !isset($_SESSION['username']) || empty($_SESSION['username'])){/* this means user is not login */ }` – Jazzzzzz Oct 23 '16 at 12:51